Number of real roots of $3x^4+6x^3+x^2+6x+3$ How many real roots does the following quartic polynomial have?
$$3x^4+6x^3+x^2+6x+3$$
   After dividing both sides by $x^2$, we get
$$3x^2+6x+1+\dfrac6x+\dfrac3{x^2}=0$$
Or,$$3\left(x^2+\dfrac1{x^2}\right)+6\left(x+\dfrac1x\right)+1=0$$
Taking $x+\dfrac1x$ as $t$
$$3t^2-2+6t+1=0$$
Or,$$3t^2+6t-1=0$$
On solving I got the roots
$$\dfrac{-3+2\sqrt6}3$$ and $$\dfrac{-3-2\sqrt6}3$$
Then I plugged in the values and found only 2 roots are real use discriminant
 A: You have committed a mistake. The correct solution after substitution is 
$$3(t^2-2)+6t+1=0\\
\implies3t^2+6t-5=0$$
Note that it takes value $3\times4+12-5>0$ at $2$ and $-5<0$ at $-2$. Hence, there is precisely one root between $2$ and $-2$. But we have $2$ solutions of the original quartic equation for each value of $|t|\ge2$. Hence, the original quartic equation has $2$ real solutions.
A: This is a program for finding out. There are at least two real roots since there is one between $-1$ and $0,$ by IVT. Then by continuity and the fact that the polynomial is positive for large values of $|x|,$ there must be another real root, clearly a negative one, since for $x>0,$ we have that the polynomial always rakes positive values.
To find out the status of the other roots, find out how the function behaves by considering its first two derivatives. The first gives $12x^3+18x^2+2x+6,$ which definitely has a real root, clearly negative. We may find out the other roots of this by again taking derivatives, which gives $36x^2+36x+2,$ which has two real roots, both negative. These roots are $$\frac{-3\pm\sqrt 7}{6}.$$ The derivative of this gives $72x+36,$ which is negative for any $x<-1/2.$ Since $\frac{-3-\sqrt 7}{6}$ is less than $-1/2,$ then it follows that the cubic has a maximum value there. At the other root $\frac{-3+\sqrt 7}{6}>-1/2,$ the cubic has a minimum value. Since the cubic is negative for large negative $x$ and positive for large positive $x,$ then the sign of either of the extreme value fixes whether the other two roots are real or not. The maximum occurs near $-1/2,$ in the interval $(-1,0).$ Here the cubic has a clearly positive value. Hence the maximum value of the cubic is positive. Therefore all its roots are real.
This means that the stationary points of the quartic are three in number. If none of them is a saddle point, then likely all the roots are real. Now since the quadratic is positive in the interval $(-\infty, r_1),$ with $r_1$ being its more negative root, then for the root of the cubic in this interval, we deduce that the quartic has a minimum here. Thus the other two roots of the cubic are in $(r_1,0).$ In $(r_1,r_2),$ $r_2$ being the less negative of the roots of the quadratic, the quadratic is negative. Thus the roots of the cubic here indicate that there is a maximum value. The last stationary point of the quartic must then necessarily be a minimum. Finally, if we show that the two minimum values of the quartic are negative (or that one minimum is negative and the maximum positive), then all its roots must be real. To do this, note that the minimum in $(-1,0)$ must be negative since the quartic is negative when $x=-1/2.$ As for the maximum value, it also occurs in $(-1,0).$ This has to be positive since there is a root here. And now the proof is complete.
A: $$x = \frac{-3 - 2 \sqrt{6} \pm \sqrt{12 \sqrt{6} - 3}}{6}$$
There are $2$ more complex roots.
